نتایج جستجو برای: graded almost semiprime

تعداد نتایج: 230368  

Journal: :Int. J. Math. Mathematical Sciences 2005
Mohammed Salem Samman

Let R be a ring with center Z(R). We write the commutator [x, y] = xy− yx, (x, y ∈ R). The following commutator identities hold: [xy,z] = x[y,z] + [x,z]y; [x, yz] = y[x,z] + [x, y]z for all x, y,z ∈ R. We recall that R is prime if aRb = (0) implies that a= 0 or b = 0; it is semiprime if aRa = (0) implies that a = 0. A prime ring is clearly a semiprime ring. A mapping f : R→ R is called centrali...

Journal: :Bulletin of the Australian Mathematical Society 1975

Journal: :Aequationes Mathematicae 2003

Journal: :Hacettepe Journal of Mathematics and Statistics 2016

Journal: :Czechoslovak Mathematical Journal 1981

Journal: :bulletin of the iranian mathematical society 2015
k. al-zoubi m. jaradat r. abu-dawwas

‎let $g$ be a group with identity $e.$ let $r$ be a $g$-graded‎ ‎commutative ring and $m$ a graded $r$-module‎. ‎in this paper‎, ‎we‎ ‎introduce several results concerning graded classical prime‎ ‎submodules‎. ‎for example‎, ‎we give a characterization of graded‎ ‎classical prime submodules‎. ‎also‎, ‎the relations between graded‎ ‎classical prime and graded prime submodules of $m$ are studied‎.‎

Journal: :iranian journal of science and technology (sciences) 2013
n. a. ozkiırisci

let  be a graded ring and  be a graded -module. we define a topology on graded prime spectrum  of the graded -module  which is analogous to that for , and investigate several properties of the topology.

Journal: :iranian journal of science and technology (sciences) 2015
s. huang

a polynomial   1 2 ( , , , ) n f x x x is called multilinear if it is homogeneous and linear in every one of its variables. in the present paper our objective is to prove the following result: let   r be a prime k-algebra over a commutative ring   k with unity and let 1 2 ( , , , ) n f x x x be a multilinear polynomial over k. suppose that   d is a nonzero derivation on r such that ...

2008

Let E pq be the vertical homology Hg(E 0 p∗) of this complex. In other words, if Z pq = ker(d v pq), B pq = im(d v p,q+1), then E pq = Z 1 pq/B 1 pq. We note that d h induces maps E pq → E p−1,q as follows. The condition dd = −dd shows that d(Z pq) ⊆ Z p−1,q and d(B pq) ⊆ B p−1,q; therefore, there is a homomorphism d̃pq : E 1 pq → E p−1,q. Let E pq be the horizontal homology Hp(E ∗q) of this com...

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